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Posted on Jan 02, 2017
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Arrange each term in each binomial in order of degree from greatest to least. The degree of a binomial is the exponent attached to the term. For example, 4x^2 is a second degree term.
Multiply each term in the binomial that is being subtracted by -1 to turn it into an addition problem. For example, the problem (8x^2 + 8) - (x^2 - 2) becomes (8x^2 + 8) + (-x^2 + 2).
Combine like terms. In the example problem, the x^2 terms are combined and the constant terms are combined, yielding (8x^2 + 8) + (-x^2 + 2) = 7x^2 + 10.
Understand the F.O.I.L. method. F.O.I.L. is an acronym standing for first, outside, inside and last. It means that you multiply the first number of the first binomial by the first number of the second, then the numbers on the outside (the first term of the first binomial by the second term of the second binomial) and so on. This ensures that both numbers in the first binomial are multiplied by both numbers in the second.
Use the F.O.I.L. method to multiply the two binomials together. For example, (3x + 4)(3x - 4) = 9x^2 +12x - 12x - 16. Notice that -12x is the product of the outside terms and -16 is the product of the last terms, 4 and -4.
Simplify. There will almost always be like terms to combine. In the example, 12x and -12x cancel out, yielding the answer 9x^2 - 16.
Use the distributive property to divide both terms in the binomial by the monomial divisor. For example, (18x^3 + 9x^2) / 3x = (18x^3 / 3x) + (9x^2 / 3x).
Understand how to divide by a term. If you are dividing a higher order term by a lower order term, you subtract the exponent. For example, y^3/y = y^2. The number part of each term is handled like any other division problem. For example, 20z / 4 = 5z.
Divide each term in the binomial by the divisor; (18x^3 / 3x) + (9x^2 / 3x) = 6x^2 + 3x.
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(a+b)2 = (a+b)(a+b) = ... ?
(a+b)2 = a2 + 2ab + b2
You can easily see why it works, in this diagram:
2. Subtract Times Subtract And what happens if you square a binomial with a minus inside?
(a-b)2 = (a-b)(a-b) = ... ?
(a-b)2 = a2 - 2ab + b2
3. Add Times Subtract And then there is one more special case... what if you multiply (a+b) by (a-b) ?
(a+b)(a-b) = ... ?
(a+b)(a-b) = a2 - b2
That was interesting! It ended up very simple.
And it is called the "difference of two squares" (the two squares are a2 and b2).
This illustration may help you see why it works:
a2 - b2 is equal to (a+b)(a-b) Note: it does not matter if (a-b) comes first:
(a-b)(a+b) = a2 - b2
The Three Cases Here are the three results we just got:
(a+b)2 = a2 + 2ab + b2 } (the "perfect square trinomials") (a-b)2 = a2 - 2ab + b2 (a+b)(a-b) = a2 - b2 (the "difference of squares") Remember those patterns, they will save you time and help you solve many algebra puzzles.
Using Them So far we have just used "a" and "b", but they could be anything.
We can use the (a+b)2 case where "a" is y, and "b" is 1:
(y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1
We can use the (a-b)2 case where "a" is 3x, and "b" is 4:
(3x-4)2 = (3x)2 - 2(3x)(4) + (4)2 = 9x2 - 24x + 16
We know that the result will be the difference of two squares, because:
(a+b)(a-b) = a2 - b2
(4y+2)(4y-2) = (4y)2 - (2)2 = 16y2 - 4
Sometimes you can recognize the pattern of the answer:
Example: can you work out which binomials to multiply to get 4x2 - 9
Hmmm... is that the difference of two squares?
Yes! 4x2 is (2x)2, and 9 is (3)2, so we have:
4x2 - 9 = (2x)2 - (3)2
And that can be produced by the difference of squares formula:
(a+b)(a-b) = a2 - b2
Like this ("a" is 2x, and "b" is 3):
(2x+3)(2x-3) = (2x)2 - (3)2 = 4x2 - 9
So the answer is that you can multiply (2x+3) and (2x-3) to get 4x2 - 9
Jul 26, 2011 | Computers & Internet
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Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:
The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.
The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.
This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.
This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y
This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2
This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3
This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4
This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5
This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6
Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.
binomial coefficient (n over k) for (x+y)^6
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